Shear viscosity and electrical conductivity of the relativistic fluid in the presence of a magnetic field: A massless case

Abstract

We have explored the shear viscosity and electrical conductivity calculations for bosonic and fermionic media, without and with presence of an external magnetic field. For numerical visualisation, we have dealt with their simplified massless expressions. In the presence of a magnetic field, five independent velocity gradient tensors can be designed, and so their corresponding proportional coefficients, connected with the viscous stress tensor, provide us five components of the shear viscosity coefficient. In the existing literature, two sets of viscous stress tensors are available. Starting from them, the present work has obtained expressions for two sets of five shear viscosity coefficients, which can be ultimately classified into three basic components – parallel, perpendicular and Hall components as one get similar expression for the electrical conductivity at the finite magnetic field. Our calculations are based on the kinetic theory approach in relaxation time approximation. Repeating the same mathematical steps under finite magnetic field, which is traditionally practiced in the absence of magnetic field, we have obtained two sets of five shear viscosity components, whose final expressions are in good agreements with earlier references, although a difference in methodology or steps can be noticed. In this context, the present work, for the first time, addresses a detailed calculation of relaxation time approximation (RTA)-based kinetic theory calculations of the second set of five shear viscosity components, which was previously done by Denicol et al (Phys. Rev. D 98, 076009 (2018)) in moment method technique. Realising the massless results of viscosity and conductivity for Maxwell–Boltzmann, Fermi–Dirac and Bose–Einstein distribution functions, we have applied them for massless quark gluon plasma and hadronic matter phases, which can provide us a rough order of strength, within which actual results will vary during quark–hadron phase transition. The present work also indicates that the magnetic field might have some role in building perfect fluid nature in RHIC or LHC matter. The lower bound expectation of shear viscosity to entropy density ratio is also discussed. Here, for the first time, we are addressing an analytic expression of temperature- and magnetic field-dependent relaxation time of the massless fluid, for which perpendicular component of shear viscosity to entropy density ratio can reach its lower bound.

Appendices

Appendix A: $C_n$ calculations

1.1 Appendix A.1. For tensors of ref. [18]

The detailed calculations of $C_n$ for traceless independent tensors of ref. [18] from the RTA-based RBE, given in eq. (10), will be documented in this section.

For zero bulk viscosity and knowing that the tensors $V_{ij}$ is symmetric, $b_{ij}$ is antisymmetric in i, j we get some relations: $\mathbf {\nabla } \cdot \mathbf {V}=V_{ii}=0$, $V_{ij}b_{i}b_{j}=0$, $b_{ij}b_{i}=b_{ij}b_{j}=0$, $b_{ij}v_{i}v_{j}=0$. Using the above conditions, eq. (10) takes the form

$$\begin{aligned}&\frac{\omega }{T} v_{i}v_{j}V_{ij}f_0(1-a f_0) = -\dfrac{2}{\tau _B}\bigg [C_1\Big \{2V_{ik}b_{ij} v_{j} v_{k}\nonumber \\&\qquad - 2V_{ik}b_{ij}b_{k}v_{j} (\vec {v} \cdot \vec {{b}})\Big \} + C_2\Big \{2V_{ik}b_{ij}v_{j}b_{k}(\vec {v} \cdot \vec {{b}}) \Big \}\nonumber \\&\qquad + C_3\Big \{2V_{ij}v_{i}v_{j} - 4V_{ij}v_{i}b_{j}(\vec {v} \cdot \vec {{b}})\Big \}\nonumber \\&\qquad + C_4\Big \{2V_{ij}v_{i}b_{j}(\vec {v} \cdot \vec {{b}})\Big \}\bigg ] + \frac{1}{\tau _c}\bigg [C_1\Big \{2V_{ij} v_i v_j\nonumber \\&\qquad - 4V_{ij}v_{i}b_{j}(\vec {v} \cdot \vec {{b}})\Big \} C_2\Big \{4V_{ij} v_i b_j(\vec {v} \cdot \vec {{b}})\Big \}\nonumber \\&\qquad + C_3\Big \{2V_{ik}b_{jk}v_i v_j - 2 V_{ij}b_{ki}b_{j}v_{k}(\vec {v} \cdot \vec {{b}})\Big \}\nonumber \\&\qquad + C_4\Big \{4V_{ij}b_{ki}b_{j} v_{k}(\vec {v} \cdot \vec {{b}})\Big \}\bigg ]~. \end{aligned}$$

(A.1)

Now, from eq. (A.2) comparing the following tensor structures,

$$\begin{aligned}&V_{ij}v_i v_j :~~ -\frac{\omega }{T}f_0 (1-af_0) = \frac{4}{\tau _B}C_3 - \frac{2}{\tau _c}C_1 \end{aligned}$$

(A.2)

$$\begin{aligned}&V_{ij}b_{ik}v_k v_j:~~ \frac{4}{\tau _B}C_1 + \frac{2}{\tau _c}C_3 = 0 \end{aligned}$$

(A.3)

$$\begin{aligned}&V_{ij}b_{ik}v_k b_j (\vec {v} \cdot \vec {{b}}):~~ \nonumber \\&\qquad \frac{2}{\tau _B}(-2C_1 + 2C_2)+ \frac{1}{\tau _c}(-2C_3 + 4C_4) = 0 \end{aligned}$$

(A.4)

$$\begin{aligned}&V_{ij}v_i b_j (\vec {v} \cdot \vec {{b}}):~~ \nonumber \\&\qquad \frac{2}{\tau _B}(-4C_3 + 2C_4) - \frac{1}{\tau _c}(-4C_1 + 4C_2) = 0. \end{aligned}$$

(A.5)

Solving the above four equations, we finally get Cs as

$$\begin{aligned} C_1= & {} -\frac{\omega }{2T}\; \frac{\tau _c}{4\left\{ \frac{1}{4}+(\tau _c/\tau _B)^2\right\} } ~ f_0(1-a f_0)\nonumber \\ C_2= & {} -\frac{\omega }{2T}\; \frac{\tau _c}{1+(\frac{\tau _c}{\tau _B})^2} ~ f_0(1-a f_0)\nonumber \\ C_3= & {} -\frac{\omega }{2T}\; \frac{\tau _c(\frac{\tau _c}{\tau _B})}{2\left\{ \frac{1}{4}+(\tau _c/\tau _B)^2\right\} } ~ f_0(1-a f_0)\nonumber \\ C_4= & {} -\frac{\omega }{2T}\; \frac{\tau _c(\frac{\tau _c}{\tau _B})}{1+(\frac{\tau _c}{\tau _B})^2} ~ f_0(1-a f_0). \end{aligned}$$

(A.6)

1.2 Appendix A.2. For tensors of refs [61, 62]

Here, $C_n$s for tensor of eq. (19) [61, 62] will be calculated.

Using $C_{ij}$s of eq. (19) in eq. (10) with tensor identities used in the above subsection we get

$$\begin{aligned}&\frac{\omega }{T} v_{i}v_{j}V_{ij}f_0(1-a f_0) = -\frac{2}{\tau _B}\bigg [C_0\left( 2V_{ik}b_{ij} v_{j} v_{k}\right) \nonumber \\&\quad C_2\Big \{2V_{ik}b_{ij}v_{j}b_{k}(\vec {v} \cdot \vec {{b}}) \Big \} \nonumber \\&\quad + C_3\Big \{4V_{ij}v_{i}v_{j} - 8V_{ij}v_{i}b_{j}(\vec {v} \cdot \vec {{b}})\Big \}\nonumber \\&\quad + C_4\Big \{2V_{ij}v_{i}b_{j}(\vec {v} \cdot \vec {{b}})\Big \}\bigg ] \nonumber \\&\quad + \frac{1}{\tau _c}\bigg [C_0\left( 2V_{ij} v_i v_j\right) + C_2\Big \{4V_{ij} v_i b_j(\vec {v} \cdot \vec {{b}})\Big \}\nonumber \\&\quad + C_3\Big \{4V_{ik}b_{jk}v_i v_j - 4 V_{ij}b_{ki}b_{j}v_{k}(\vec {v} \cdot \vec {{b}})\Big \} \nonumber \\&\quad + C_4\Big \{4V_{ij}b_{ki}b_{j} v_{k}(\vec {v} \cdot \vec {{b}})\Big \}\bigg ]. \end{aligned}$$

(A.7)

Now, from eq. (A.7) comparing the following tensor structures,

$$\begin{aligned}&V_{ij}v_i v_j:~~ -\frac{\omega }{T}f_0 (1-af_0) = \frac{8}{\tau _B}C_3 - \frac{2}{\tau _c}C_0 \end{aligned}$$

(A.8)

$$\begin{aligned}&V_{ij}b_{ik}v_k v_j:~~ \frac{4}{\tau _B}C_0 + \frac{4}{\tau _c}C_3 = 0 \end{aligned}$$

(A.9)

$$\begin{aligned}&V_{ij}b_{ik}v_k b_j (\vec {v} \cdot \vec {{b}}): \nonumber \\&\quad \frac{2}{\tau _B}(2C_2) + \frac{1}{\tau _c}(-4C_3 + 4C_4) = 0 \end{aligned}$$

(A.10)

$$\begin{aligned}&V_{ij}v_i b_j (\vec {v} \cdot \vec {{b}}): \nonumber \\&\quad \frac{2}{\tau _B}(-8C_3 + 2C_4) - \frac{1}{\tau _c}(4C_2) = 0. \end{aligned}$$

(A.11)

Solving the above four equations, we finally get Cs as

$$\begin{aligned}&C_0 = -\frac{\omega }{2T}\; \frac{\tau _c}{\left\{ 1+4(\tau _c/\tau _B)^2\right\} } ~ f_0(1-a f_0) \nonumber \\&C_2 = -\frac{\omega }{2T}\; \frac{3\tau _c(\tau _c/\tau _B)^2}{\left\{ 1+(\tau _c/\tau _B)^2\right\} \left\{ 1+4(\tau _c/\tau _B)^2\right\} } \nonumber \\&\quad f_0(1-a f_0) \nonumber \\&C_3 = -\frac{\omega }{2T}\; \frac{\tau _c(\tau _c/\tau _B)}{\left\{ 1+4(\tau _c/\tau _B)^2\right\} } ~ f_0(1-a f_0)\nonumber \\&C_4 = -\frac{\omega }{2T}\; \frac{\tau _c(\tau _c/\tau _B)}{1+(\tau _c/\tau _B)^2} ~ f_0(1-a f_0). \end{aligned}$$

(A.12)

Appendix B: Massless results

1.1 Appendix B.1. Thermodynamics of massless quark at $B=0$

Here we have shown explicit calculation of energy density, pressure and entropy density for quasiparticle system considering mass $m=0$ for MB, BE, FD at $B=0$.

Thermal distribution function can be written in a general way in the form

$$\begin{aligned} f = \frac{1}{e^{\beta E} + a}, \end{aligned}$$

(B.1)

where $a=0$ for MB, $a=1$ for FD and $a=-1$ for BE statistics.

BE: Energy density for bosons is given by

$$\begin{aligned} \epsilon = g\int _0^{\infty }\frac{\mathrm {d}^3p}{(2\pi )^3}\frac{E}{\mathrm {e}^{\beta E}-1}. \end{aligned}$$

(B.2)

Here $E = p$ which gives us

$$\begin{aligned} \mathrm {d}p = \mathrm {d}E \end{aligned}$$

and the integral becomes

$$\begin{aligned} \epsilon= & {} g\int _0^{\infty } \dfrac{4\pi E^3\mathrm {d}E}{(2\pi )^3}\dfrac{1}{\mathrm {e}^{\beta E} -1}\nonumber \\= & {} g\int _0^{\infty }\dfrac{E^3\mathrm {d}E}{2\pi ^2}\dfrac{1}{{\mathrm{e}}^{\beta E} -1}. \end{aligned}$$

(B.3)

Substituting $\beta E = x$ giving

$$\begin{aligned} \mathrm {d}E = \frac{\mathrm {d}x}{\beta } \end{aligned}$$

gives us

$$\begin{aligned} \epsilon = \frac{g_b}{2\pi ^2\beta ^4} \int _0^{\infty }\frac{x^3\mathrm {d}x}{\mathrm {e}^x-1}. \end{aligned}$$

(B.4)

This integral can be converted into a $\zeta (s)$ function by using

$$\begin{aligned} \zeta (s) = \frac{1}{\Gamma (s)}\int _0^{\infty }\frac{x^{s-1}\mathrm {d}x}{\mathrm{e}^x-1}, \end{aligned}$$

(B.5)

where $\Gamma (s)$ is the gamma function.

$$\begin{aligned} \epsilon= & {} g \times \frac{(k_BT)^4}{2\pi ^2}\zeta (4)\Gamma (4)\nonumber \\= & {} g\frac{\pi ^2}{30}T^4. \end{aligned}$$

(B.6)

Using $\zeta (4) = {\pi ^4}/{90}$ and $\Gamma (4) = 3!$, the corresponding pressure density is

$$\begin{aligned} P = \frac{\epsilon }{3} = gT^4\frac{\pi ^2}{90}. \end{aligned}$$

(B.7)

The entropy density is

$$\begin{aligned} s = \frac{\epsilon + P}{T} = \frac{4g\pi ^2}{90}T^3. \end{aligned}$$

FD: Energy density for fermion is

$$\begin{aligned} \epsilon = g \int \frac{\mathrm {d}^3p}{(2\pi )^3}E \frac{1}{\mathrm{e}^{\beta E} + 1}. \end{aligned}$$

(B.8)

$E = p$ which gives the integral

$$\begin{aligned} \epsilon = \frac{g}{2\pi ^2}\int _0^{\infty } \frac{E^3 {d}E}{\mathrm{e}^{\beta E} + 1}. \end{aligned}$$

(B.9)

Substituting $\beta E = x$ we get

$$\begin{aligned} \mathrm {d}E = \frac{\mathrm {d}x}{\beta } \end{aligned}$$

$$\begin{aligned} \epsilon= & {} \frac{g}{2\pi ^2}T^4\Gamma (4)\left( 1-\frac{1}{2^{4-1}}\right) \zeta (4)\nonumber \\= & {} \frac{7}{8}g\frac{\pi ^2}{30}T^4. \end{aligned}$$

(B.10)

Pressure density of fermions is given by

$$\begin{aligned} P = \frac{\epsilon }{3} = \frac{7}{8}g\frac{\pi ^2}{90}T^4. \end{aligned}$$

(B.11)

The entropy density is given by

$$\begin{aligned} s = \frac{\epsilon + P}{T} = \frac{7g\pi ^2}{90}T^3. \end{aligned}$$

MB: For the Maxwell–Boltzmann distribution, the energy density is given by

$$\begin{aligned} \epsilon = g \int _0^{\infty } \frac{\mathrm{d}^3p }{(2\pi )^3}\frac{E}{\mathrm{e}^{\beta E}} \end{aligned}$$

(B.12)

Since $E = p$ the energy density calculated by changing the integral to a beta function is

$$\begin{aligned} \epsilon = \frac{3gT^4}{\pi ^2}. \end{aligned}$$

(B.13)

The pressure density is $P = {\epsilon }/{3}$

$$\begin{aligned} P = \frac{g}{\pi ^2}T^4. \end{aligned}$$

(B.14)

The entropy density is given by

$$\begin{aligned} s = \frac{\epsilon + P }{T} = \frac{4g}{\pi ^2}T^3. \end{aligned}$$

(B.15)

1.2 Appendix B.2. Shear viscosity and electrical conductivity of massless quark at $B=0$

Here we have shown the detailed derivation of $\eta (T)$ and $\sigma (T)$ for MB, BE, FD at $B= 0$ for massless quasiparticle system. The shear viscosity $\eta$ for bosons is given by

$$\begin{aligned} \eta = \frac{g\beta }{15} \int \frac{\mathrm {d}^3p}{(2\pi )^3}\frac{p^4}{E^2}\tau _cf_0(1+f_0). \end{aligned}$$

(B.16)

Here $E = p$ which gives us

$$\begin{aligned} \eta = \frac{g\beta }{15}\frac{\tau _c}{2\pi ^2}\int _0^{\infty }p^4 \mathrm {d}pf_0(1+f_0). \end{aligned}$$

(B.17)

The shear viscosity for fermions is given by

$$\begin{aligned} \eta= & {} \frac{g\beta }{15}\int \frac{4\pi p^2 \mathrm {d}p }{(2\pi )^3}p^2\tau _cf_0(1-f_0)\nonumber \\= & {} \frac{g\beta }{15}\frac{\tau _c}{2\pi ^2}\int _0^{\infty }p^4\mathrm {d}pf_0(1-f_0). \end{aligned}$$

(B.18)

The above two expressions are written as

$$\begin{aligned} \eta = A \int _0^{\infty }p^4\mathrm {d}pf_0(1-f_0) = A I_1 \end{aligned}$$

(B.19)

for fermions and

$$\begin{aligned} \eta = A \int _0^{\infty } p^4\mathrm {d}p f_0(1+f_0) = AI_2 \end{aligned}$$

(B.20)

for bosons, where the constant A is the substitution for

$$\begin{aligned} A = \frac{g\beta }{30\pi ^2}\tau _c. \end{aligned}$$

(B.21)

$I_1$ integral is evaluated as follows:

$$\begin{aligned} I = \int _0^{\infty } p^4 \mathrm {d}p f_0(1-f_0), \end{aligned}$$

(B.22)

where

$$\begin{aligned} f_0 = \frac{1}{\mathrm{e}^{\beta E}+1} = \frac{1}{\mathrm{e}^{\beta p}+1} \end{aligned}$$

is the distribution function for fermions.

$$\begin{aligned} I_1= & {} \int _0^{\infty } p^4 \mathrm {d}p \frac{1}{\mathrm{e}^{\beta p}+1}\left[ 1- \frac{1}{\mathrm{e}^{\beta p}+1}\right] \nonumber \\= & {} \int _0^{\infty } p^4 \mathrm {d}p \frac{\mathrm{e}^{\beta p}}{(\mathrm{e}^{\beta p}+1)^2}\nonumber \\= & {} -\frac{\partial }{\partial \beta }\int _0^{\infty } \frac{p^3}{\mathrm{e}^{\beta p} +1} \mathrm {d}p. \end{aligned}$$

(B.23)

By using the definition of d(s) and $\zeta (s)$ functions we solve the above integral as

$$\begin{aligned} I_1= & {} -\frac{\partial }{\partial \beta }\frac{\Gamma (4)}{(\beta )^4}\left( 1-\frac{1}{2^3}\right) \zeta (4) = -\frac{\partial }{\partial \beta }\frac{3!}{\beta ^4}\frac{7}{8}\zeta (4)\nonumber \\= & {} \frac{4!}{\beta ^5}\zeta (4)\frac{7}{8}. \end{aligned}$$

Thus

$$\begin{aligned} \eta _2|_{\mathrm {fermions}} = A \frac{4!}{\beta ^5}\zeta (4)\frac{7}{8}. \end{aligned}$$

The integral for bosons is

$$\begin{aligned} I_2= & {} \int _0^{\infty }p^4\mathrm {d}p\Big (\frac{1}{\mathrm {e}^{\beta p} -1}\Big )\Big (1+\frac{1}{\mathrm {e}^{\beta p}-1}\Big )\nonumber \\= & {} \int _0^{\infty }p^4 \mathrm {d}p \frac{\mathrm {e}^{\beta p}}{(\mathrm {e}^{\beta p}-1)^2}\nonumber \\= & {} -\frac{\partial }{\partial \beta }\int _0^{\infty }p^3\mathrm {d}p \frac{\mathrm {e}^{\beta p}}{\mathrm {e}^{\beta p}-1}. \end{aligned}$$

(B.24)

Substituting $\beta p = x$ we get

$$\begin{aligned} I_2 = -\frac{\partial }{\partial \beta }\Big (\int _0^{\infty }\mathrm {d}x\frac{x^3}{\beta ^4(\mathrm {e}^x-1)} \Big ). \end{aligned}$$

(B.25)

Using the definition of $\zeta (s)$ the integral is calculated as

$$\begin{aligned} I_2 = \frac{4}{\beta ^5}\Gamma (4)\zeta (4). \end{aligned}$$

(B.26)

$\eta |_{\mathrm {bosons}}$ is obtained as

$$\begin{aligned} \eta |_{\mathrm {bosons}} = A \frac{4}{\beta ^5}\Gamma (4)\zeta (4). \end{aligned}$$

(B.27)

For Maxwell–Boltzmann distribution, the shear viscosity is obtained as follows:

$$\begin{aligned} \eta = \frac{g\beta }{15}\int \frac{\mathrm {d}^3p}{(2\pi )^3}p^2f_0 \tau _c, \end{aligned}$$

(B.28)

where $f_0 = \mathrm {e}^{-\beta E}$ and here $E = p$.

$$\begin{aligned} \eta= & {} A \int _0^{\infty } p^4 \mathrm {d}p \mathrm {e}^{-\beta E} = \frac{A}{\beta ^5}\int _0^{\infty } \mathrm {e}^{-x}x^4 \mathrm {d}x\nonumber \\= & {} \frac{A}{\beta ^5}\Gamma (5), \end{aligned}$$

(B.29)

where

$$\begin{aligned} A = \frac{g\beta }{30\pi ^2}\tau _c \end{aligned}$$

and

$$\begin{aligned} \Gamma (5) = 4! \end{aligned}$$

Electrical conductivity $\sigma$ for different distributions is calculated as follows:

MB: For Maxwell–Boltzmann distribution, the electrical conductivity is

$$\begin{aligned} \sigma= & {} \frac{q^2 g \beta }{3}\int \frac{\mathrm {d}^3p}{(2\pi )^3}\frac{p^2}{E^2}\tau _cf_0\nonumber \\= & {} \frac{q^2 g \beta }{3}\int \frac{4\pi p^2 \mathrm {d}p}{(2\pi )^3}\tau _cf_0. \end{aligned}$$

(B.30)

The distribution function is $f_0 = \mathrm {e}^{-\beta p}$ for Maxwell–Boltzmann distribution.

$$\begin{aligned} \sigma = \frac{q^2 g \beta }{3(2\pi ^2)}\int _0^{\infty } p^2 \mathrm {d}p [\tau _c \mathrm {e}^{-\beta p}]. \end{aligned}$$

(B.31)

Substituting $\beta p = x$ we get

$$\begin{aligned} \sigma= & {} \frac{q^2 g \beta }{3(2\pi ^2)}\tau _c\int _0^{\infty } \frac{1}{\beta ^3} \mathrm {d}x x^2 \mathrm {e}^{-x} = \frac{q^2 g \beta }{3(2\pi ^2)} \frac{2!}{\beta ^3}\tau _c\nonumber \\= & {} q^2\frac{g}{3\pi ^2}\frac{\tau _c}{\beta ^2}. \end{aligned}$$

(B.32)

FD: The electrical conductivity of fermions is given by

$$\begin{aligned} \sigma = q^2\frac{g\beta }{3}\int \frac{\mathrm {d}^3p}{(2\pi )^3}\frac{p^2}{E^2}\tau _cf_0(1-f_0), \end{aligned}$$

(B.33)

where the distribution function of fermions is given by

$$\begin{aligned} f_0 = \frac{1}{\mathrm {e}^{\beta p} +1} \end{aligned}$$

$$\begin{aligned} \sigma= & {} q^2\frac{g\beta }{3} \int _0^{\infty } \frac{4\pi p^2 }{(2\pi )^3}\mathrm {d}p \times \tau _c\frac{\mathrm {e}^{\beta p}}{(\mathrm {e}^{\beta p} +1)^2}\nonumber \\= & {} \frac{q^2 g \beta }{3\times 2 \pi ^2}\int _0^{\infty }p^2 \mathrm {d}p \frac{\mathrm {e}^{\beta p}}{(\mathrm{e}^{\beta p}+1)^2}\tau _c. \end{aligned}$$

(B.34)

Using the definition of $\zeta (s)$ the integral is evaluated to be

$$\begin{aligned} \sigma = \frac{q^2 g}{2\times 2\pi ^2}\frac{1}{\beta ^2}\frac{\Gamma (4)}{4}\zeta (3)\tau _c, \end{aligned}$$

(B.35)

where $\Gamma (4) =3!$

BE: For bosons, $\sigma$ is

$$\begin{aligned} \sigma = \frac{q^2g\beta }{3}\int \frac{\mathrm {d}^3p}{(2\pi )^3}\frac{p^2}{E^2}f_0(1+f_0)\tau _c, \end{aligned}$$

(B.36)

where the distribution function of bosons is given by

$$\begin{aligned} f_0 = \frac{1}{\mathrm {e}^{\beta p} -1} \end{aligned}$$

$$\begin{aligned} \sigma= & {} \frac{q^2g\beta }{3\times 2\pi ^2}\int _0^{\infty } \frac{p^2 \mathrm {e}^{\beta p}}{(\mathrm {e}^{\beta p}-1)^2}\mathrm {d}p \tau _c\nonumber \\= & {} \frac{q^2g\beta }{3\times 2\pi ^2} \times I \times \tau _c, \end{aligned}$$

(B.37)

where the integral I is given by

$$\begin{aligned} I = \int _0 ^{\infty } \frac{p^2 \mathrm {e}^{\beta p}}{(\mathrm {e}^{\beta p} -1)^2}\mathrm {d}p \end{aligned}$$

(B.38)

is solved as follows:

$$\begin{aligned} I= & {} -\frac{\partial }{\partial \beta }\int _0^{\infty } \frac{p}{\mathrm {e}^{\beta p}-1}\mathrm {d}p\nonumber \\= & {} -\frac{\partial }{\partial \beta }\Big (\frac{1}{\beta ^2}\Big )\int _0^{\infty } \frac{x}{\mathrm {e}^x-1}\mathrm {d}x \nonumber \\= & {} \frac{2}{\beta ^3}\int _0^{\infty }\frac{x}{\mathrm {e}^x-1}\mathrm {d}x\nonumber \\= & {} \frac{2}{\beta ^3}\zeta (2)\Gamma (2) = \frac{2}{\beta ^3}\zeta (2). \end{aligned}$$

(B.39)

Putting this in the expression for conductivity, we get

$$\begin{aligned} \sigma = \frac{q^2 g \beta }{3\times 2 \pi ^2}\frac{2}{\beta ^3}\zeta (2)\tau _c. \end{aligned}$$

(B.40)

1.3 Appendix B.3. Thermal average energy

The expressions of viscosity and conductivity contain magnetic relaxation time $\tau _B$ which is $\tau _B={E}/{qB}$. But for simplicity of calculation we shall consider average energy for $\tau _B$ calculation. So, $\tau _B={\langle E \rangle }/{qB}.$

MB: Average energy for Maxwell–Boltzmann distribution with $E =p$

$$\begin{aligned} \langle E \rangle= & {} \frac{\int \frac{\mathrm {d}^3p}{(2\pi )^3}E \mathrm {e}^{-\beta E}}{\int \frac{\mathrm {d}^3p}{(2\pi )^3}\mathrm {e}^{-\beta E}} = \frac{\int _0^{\infty }p^3\mathrm {d}p\mathrm {e}^{-\beta p}}{\int p^2\mathrm {e}^{-\beta p}\mathrm {d}p}\nonumber \\= & {} \frac{\frac{6}{\beta ^4}}{\frac{2}{\beta ^3}} = \frac{3}{\beta }. \end{aligned}$$

(B.41)

FD: Average energy for fermions is

$$\begin{aligned} \langle E \rangle = \frac{\int \frac{\mathrm {d}^3p}{(2\pi )^3}E \frac{1}{\mathrm {e}^{\beta E}+1}}{\int \frac{\mathrm {d}^3p}{(2\pi )^3}\frac{1}{\mathrm {e}^{\beta E}+1}} = \frac{\int _0^{\infty } \frac{p^3 \mathrm {d}p}{\mathrm {e}^{\beta p}+1}}{\int _0^{\infty }\frac{p^2\mathrm {d}p}{\mathrm {e}^{\beta p}+1}}. \end{aligned}$$

(B.42)

Using the definition of $\zeta (s)$ and substituting $\beta p =x$ the above integral is evaluated as

$$\begin{aligned} \langle E \rangle = \frac{\frac{\Gamma (4)}{\beta ^4}\Big (1-\frac{1}{2^{4-1}}\Big )\zeta (4)}{\frac{\Gamma (3)}{\beta ^3}\Big (1-\frac{1}{2^{3-1}}\Big )\zeta (3)} = \frac{7}{2}T\frac{\zeta (4)}{\zeta (3)}. \end{aligned}$$

(B.43)

BE: Average energy of bosons with $E =p$

$$\begin{aligned} \langle E \rangle = \frac{\int \frac{\mathrm {d}^3p}{(2\pi )^3}\frac{E}{(\mathrm {e}^{\beta E}-1)}}{\int \frac{\mathrm {d}^3p}{(2\pi )^3}\frac{1}{(\mathrm {e}^{\beta E}-1)}} = \frac{\int _0^{\infty }\frac{p^3 \mathrm {d}p}{(\mathrm {e}^{\beta p}-1)}}{\int _0^{\infty }\frac{p^2 \mathrm {d}p}{(\mathrm {e}^{\beta p}-1)}}. \end{aligned}$$

(B.44)

Using the definition of $\zeta (s)$ and substituting $\beta p =x$ the integral can be solved as

$$\begin{aligned} \langle E \rangle = \frac{1}{\beta } \frac{\zeta (4)\Gamma (4)}{\zeta (3)\Gamma (3)}. \end{aligned}$$

(B.45)